1. I need the complete proof for commutation relation of the Lorentz group generators.

  2. The proof of Lorentz algebra using this commutation relation.


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The Lorentz group is the group of matrices that conserve the quadratic form:

$$\mathscr{Q}(X,\,Y) = X^T\,\eta\,Y\tag{1}$$

where here $X$ and $Y$ are $1\times 4$ column vectors, the $4\times 4$ group member matrices act on these from the left and $\eta$ is the Minkowski (pseuso) metric. Therefore, $\Lambda\in O(1,\,3)$ if and only if:

$$\mathscr{Q}(\Lambda\,X,\,\Lambda\,Y) = X^T\,\Lambda^T\,\eta\,\Lambda\,Y=X^T\,\eta\,Y;\;\forall\,X,\,Y\in\mathbb{R}^{1+3}\tag{2}$$

Exercise 1: Show that (2) can only hold for for any $\Lambda(s) = \exp(s\, L)\in O(1,\,3)\,\forall\,s\in\mathbb{R}$ for a given member $L$ of the Lie algebra of $O(1,\,3)$ (here written as a $4\times 4$ matrix) if and only if:

$$L^T\,\eta + \eta\,L = 0\tag{3}$$

This then is the defining relationship you must use for the Lie algebra members.

Exercise 2: Show that $L$ fulfills (3) if and only if $L$ is of the form:

$$L = \left(\begin{array}{cccc}0&\eta_x&\eta_y&\eta_z\\\eta_x&0&-\theta_z&\theta_y\\\eta_y&\theta_z&0&-\theta_x\\\eta_z&-\theta_y&\theta_x&0\end{array}\right)\tag{4}$$

where all the $\eta_j,\theta_j\in\mathbb{R}$. The Mathematica code:

In[1]:= L={{L11,L12,L13,L14},{L21,L22,L23,L24},{L31,L32,L33,L44},{L41,L42,L43,L44}};
In[2]:= Solve[Transpose[L].DiagonalMatrix[{1,-1,-1,-1}]+DiagonalMatrix[{1,-1,-1,-1}].L==0,Flatten[L]]
Out[2]= {{L11->0,L21->L12,L22->0,L31->L13,L32->-L23,L33->0,L44->0,L41->L14,L42->-L24,L43->0}}

will help here.

Thus you can write a general Lie algebra member as the following superposition of the six linearly independent matrices:

$$L = \sum\limits_{i\in\{x,\,y,\,z\}}(\eta_i\, K_i +\theta_i\,J_i)\tag{5}$$

where the $J_i$ are the "infinitessimal" rotations and span the Lie algebra of the rotation group $SO(3)$ and the $K_i$ are the "infinitessimal" boosts.

Exercise 3: Find the required commutation relationships using the definitions of $J_i,\,K_i$ that follow from (5).

An interesting aside:

Exercise: Using bilinearity of the form in (1), and considering each of the entities $\mathscr{Q}(X,\,X)$, $\mathscr{Q}(Y,\,Y)$, ,$\mathscr{Q}(X+Y,\,X+Y)$ and $\mathscr{Q}(X,\,Y)$ show that the condition (2) holds if and only if the seemingly weaker condition:

$$\mathscr{Q}(\Lambda\,X,\,\Lambda\,X) = X^T\,\Lambda^T\,\eta\,\Lambda\,X=X^T\,\eta\,X;\;\forall\,X\in\mathbb{R}^{1+3}\tag{6}$$



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